theorem
  A? ^^ (A*) = A*
proof
  thus A? ^^ (A*) = ({<%>E} \/ A) ^^ (A*) by Th76
    .= {<%>E} ^^ (A*) \/ A ^^ (A*) by FLANG_1:20
    .= A* \/ A ^^ (A*) by FLANG_1:13
    .= A* by FLANG_1:53,XBOOLE_1:12;
end;
